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Question: Calculate the amount of heat required to convert \(5{\text{kg}}\) of ice at \(0^\circ {\text{C}}\) t...

Calculate the amount of heat required to convert 5kg5{\text{kg}} of ice at 0∘C0^\circ {\text{C}} to vapour at 100∘C100^\circ {\text{C}} .
A) 1⋅5×107J1 \cdot 5 \times {10^7}{\text{J}}
B) 2⋅5×107J2 \cdot 5 \times {10^7}{\text{J}}
C) 3⋅5×107J3 \cdot 5 \times {10^7}{\text{J}}
D) 4⋅5×107J4 \cdot 5 \times {10^7}{\text{J}}

Explanation

Solution

In the given problem, we have to convert the given amount of ice at 0∘C0^\circ {\text{C}} to water vapour at 100∘C100^\circ {\text{C}} . The process involves three steps - the conversion of ice at 0∘C0^\circ {\text{C}} to water at 0∘C0^\circ {\text{C}} , raising the temperature of water from 0∘C0^\circ {\text{C}} to 100∘C100^\circ {\text{C}} and finally the conversion of water at 100∘C100^\circ {\text{C}} to vapour at 100∘C100^\circ {\text{C}} . The total heat required will be the sum of the heat required to complete these three steps.

Formulas used:
-The heat required to convert ice to water is given by, Q=mLfQ = m{L_f} where mm is the mass of ice and Lf{L_f} is the latent heat of fusion.
-The heat required to convert water to vapour is given by, Q=mLvQ = m{L_v} where mm is the mass of water and Lv{L_v} is the latent heat of vaporization.
-The heat required to obtain a change in temperature of a substance is given by, Q=ms(Tf−Ti)Q = ms\left( {{T_f} - {T_i}} \right) where mm is the mass of the substance, ss is the specific heat capacity of the substance, Tf{T_f} is the final temperature to be attained by the substance and Ti{T_i} is its initial temperature.

Complete step by step solution.
Step 1: List the known parameters involved in the problem at hand.
The mass of ice which is to be converted is given to be m=5kgm = 5{\text{kg}} .
The initial temperature of the ice is given to be Ti=0∘C{T_i} = 0^\circ {\text{C}} .
The final temperature of the water vapour is given to be Tf=100∘C{T_f} = 100^\circ {\text{C}} .
The latent heat of fusion for ice is known to be Lf=3⋅34×105Jkg−1{L_f} = 3 \cdot 34 \times {10^5}{\text{Jk}}{{\text{g}}^{ - 1}} .
The latent heat of vaporization for water is known to be Lv=2⋅24×106Jkg−1{L_v} = 2 \cdot 24 \times {10^6}{\text{Jk}}{{\text{g}}^{ - 1}} .
The specific heat capacity of water is known to be s=4200Jkg−1∘C−1s = 4200{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}} .
Here the first step is converting ice to water at 0∘C0^\circ {\text{C}} . The second step is increasing the temperature of water from Ti=0∘C{T_i} = 0^\circ {\text{C}} to Tf=100∘C{T_f} = 100^\circ {\text{C}} . The final step is to convert water at 100∘C100^\circ {\text{C}} to vapour at the same temperature.

Step 2: Express the heat required to convert ice to water at 0∘C0^\circ {\text{C}} .
The heat required to convert ice to water at 0∘C0^\circ {\text{C}} can be expressed as Q1=mLf{Q_1} = m{L_f} ------- (1)
Substituting for m=5kgm = 5{\text{kg}} and Lf=3⋅34×105Jkg−1{L_f} = 3 \cdot 34 \times {10^5}{\text{Jk}}{{\text{g}}^{ - 1}} in equation (1) we get, Q1=5×3⋅34×105=16⋅7×105J{Q_1} = 5 \times 3 \cdot 34 \times {10^5} = 16 \cdot 7 \times {10^5}{\text{J}}
Thus the heat required for the first step of the process is obtained to be Q1=16⋅7×105J{Q_1} = 16 \cdot 7 \times {10^5}{\text{J}} .

Step 3: Express the heat required to raise the temperature of the water.
The heat required to increase the temperature of the water can be expressed as
Q2=ms(Tf−Ti){Q_2} = ms\left( {{T_f} - {T_i}} \right) --------- (2)
Substituting for s=4200Jkg−1∘C−1s = 4200{\text{Jk}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}} , m=5kgm = 5{\text{kg}} , Ti=0∘C{T_i} = 0^\circ {\text{C}} and Tf=100∘C{T_f} = 100^\circ {\text{C}} in equation (2) we get, Q2=5×4200×(100−0)=21×105J{Q_2} = 5 \times 4200 \times \left( {100 - 0} \right) = 21 \times {10^5}{\text{J}}
Thus the heat required for the second step of the process is obtained to be Q2=21×105J{Q_2} = 21 \times {10^5}{\text{J}} .

Step 4: Express the heat required to convert water at 100∘C100^\circ {\text{C}} to vapour at the same temperature and obtain the sum of the heat required for the three steps.
The heat required to convert water to vapour at 100∘C100^\circ {\text{C}} can be expressed as Q3=mLv{Q_3} = m{L_v} ------ (3)
Substituting for m=5kgm = 5{\text{kg}} and Lv=2⋅24×106Jkg−1{L_v} = 2 \cdot 24 \times {10^6}{\text{Jk}}{{\text{g}}^{ - 1}} in equation (3) we get, Q3=5×2⋅24×106=112×105J{Q_3} = 5 \times 2 \cdot 24 \times {10^6} = 112 \times {10^5}{\text{J}}
Thus the heat required for the final step of the process is obtained to be Q3=112×105J{Q_3} = 112 \times {10^5}{\text{J}} .
Now the total heat required to obtain the required conversion of ice to vapour will be expressed as Q=Q1+Q2+Q3Q = {Q_1} + {Q_2} + {Q_3} -------- (4)
Substituting for Q1=16⋅7×105J{Q_1} = 16 \cdot 7 \times {10^5}{\text{J}} , Q2=21×105J{Q_2} = 21 \times {10^5}{\text{J}} and Q3=112×105J{Q_3} = 112 \times {10^5}{\text{J}} in equation (4) we get, Q=(16⋅7×105)+(21×105)+(112×105)=149⋅7×105J≅1⋅5×107JQ = \left( {16 \cdot 7 \times {{10}^5}} \right) + \left( {21 \times {{10}^5}} \right) + \left( {112 \times {{10}^5}} \right) = 149 \cdot 7 \times {10^5}{\text{J}} \cong {\text{1}} \cdot {\text{5}} \times {\text{1}}{{\text{0}}^7}{\text{J}}
∴\therefore the heat required for the conversion is obtained to be Q≅1⋅5×107JQ \cong {\text{1}} \cdot {\text{5}} \times {\text{1}}{{\text{0}}^7}{\text{J}} .

Hence the correct option is A.

Note: The first step involves a phase change from solid (ice) to liquid (water) and the final step involves a phase change from liquid (water) to gas (vapour). In these two steps, the heat required does not involve a change in temperature. The second step, however, requires water to remain as such but with a rise in its temperature, so here the change in temperature becomes relevant. The initial temperature of the given ice is equal to the freezing point of water and the final temperature of the vapour is equal to the boiling point of water.